The gage pressure in a liquid at a depth of 3 m is read to be 42 kPa. Determine the gage pressure in the same liquid at a depth of 9 m.

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Donald Cheek · Accepted Answer

Given that the gage pressure at depth of 3 m is 42 kPa. To determine the gage pressure at depth 9 m we can consider that the pressure of the first case P1 is the gage pressure at depth of h1=3m and similarly we can define P2 as the gage pressure at depth of h2=9m. Then we could obtain P1 and P2 as: P1=ρL g h1 (1) P2=ρL g h2 (2) Dividing equ. (2) by equ. (1) P2P1=ρLgh2ρLgh1 ∴P2=P1h2h1 =42(kPa)⋅93 =126kPa

encolatgehu · Accepted Answer

Given,  Depth 1=h1=3m  Gage pressure at depth 1=Pg1=42kPa  Depth 2=h2=9m  Total pressure in liquid is given by,   P=Pa+ρgh  Where,  Pa= Atmoshpheric pressure  ρ= Density of liquid  g= Gravitational acceleration  h= depth  Gage pressure is given by,  Pg=P−Pa=ρgh  Step 2  Gage pressure at depth 1 is given by,  Pg1=ρgh1(eq.i)  Gage pressure at depth 2 is given by,  Pg2=ρgh2(eq.ii)  Taking ration of equaion (i) and (ii),  ∴Pg1Pg2=ρgh1ρgh2  ∴PgqPg2=h1h2  ∴39  ∴Pg2=42×93  ∴Pg2=126kPa  Answer:  Gage pressure at depth of 9m=126kPa

Vasquez · Accepted Answer

Answer:126 kPaExplanation:P=ρ·g·hwhere:- P is the pressure- ρ is the density of the liquid- g is the acceleration due to gravity- h is the depth of the liquidIn this problem, we are given the depth h1=3m and the gauge pressure P1=42kPa. We need to find the gauge pressure P2 at a depth h2=9m.To find P2, we can set up a ratio between the two pressures at different depths:P2P1=h2h1Substituting the given values, we have:P242kPa=9m3mSimplifying the equation gives:P2=42kPa×9m3mNow we can calculate the value of P2:P2=42kPa×3=126kPaTherefore, the gauge pressure in the liquid at a depth of 9 m is 126 kPa.

RizerMix · Accepted Answer

Step 1:Given:P=ρ·g·h where ρ is the density of the liquid, g is the acceleration due to gravity, and h is the depth of the liquid.We are given that the gage pressure at a depth of 3 m is 42 kPa. Gage pressure is the pressure measured relative to atmospheric pressure. In other words, it is the excess pressure above atmospheric pressure. Therefore, we need to convert the given gage pressure to absolute pressure by adding the atmospheric pressure.Let&#039;s assume the atmospheric pressure is Patm.At a depth of 3 m, the absolute pressure P1 is given by:P1=Patm+gage&amp;nbsp;pressure=Patm+42kPaStep 2:Now, we need to find the gage pressure at a depth of 9 m, denoted by P2.Using the hydrostatic pressure formula, we can write:P2=Patm+ρ·g·hComparing this equation with the equation for P1, we can see that P2 is simply P1 plus the additional pressure due to the increased depth of 6 m.Therefore, we can write:P2=P1+ρ·g·Δhwhere Δh is the change in depth, which is 9m−3m=6m.Substituting the known values, we have:P2=(Patm+42kPa)+(ρ·g·6m)Since we are not given the density of the liquid or the atmospheric pressure, we cannot determine the exact value of P2 without additional information. However, we can express the solution in terms of variables and known values:P2=Patm+42kPa+ρ·g·6mNote that the gage pressure at a depth of 9 m will be greater than the gage pressure at a depth of 3 m due to the increased depth.

nick1337 · Accepted Answer

To solve the problem, we can use the equation for the gage pressure in a liquid:P=ρ·g·hwhere:P is the gage pressure,ρ is the density of the liquid,g is the acceleration due to gravity, andh is the depth of the liquid.Given that the gage pressure at a depth of 3 m is 42 kPa, we can use this information to find the value of ρ·g.At a depth of 3 m:P1=ρ·g·3⇒ρ·g=P13Substituting the given values, we have:P13=42kPaNow, we can determine the gage pressure at a depth of 9 m.At a depth of 9 m:P2=ρ·g·9We can substitute the value of ρ·g from the previous equation:P2=(P13)·9⇒P2=3P1Therefore, the gage pressure at a depth of 9 m is three times the gage pressure at a depth of 3 m.

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